PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE
Nouvelle série, tome 99(113) (2016), 43–49
DOI: 10.2298/PIM151030032J
ON KNASTER’S PROBLEM
Marija Jelić
Abstract. Dold’s theorem gives sufficient conditions for proving that there is
no G-equivariant mapping between two spaces. We prove a generalization of
Dold’s theorem, which requires triviality of homology with some coefficients,
up to dimension n, instead of n-connectedness. Then we apply it to a special
case of Knaster’s famous problem, and obtain a new proof of a result of C. T.
Yang, which is much shorter and simpler than previous proofs. Also, we obtain
a positive answer to some other cases of Knaster’s problem, and improve a
result of V. V. Makeev, by weakening the conditions.
1. Introduction
Many problems in topological combinatorics can be reduced to a question about
the existence of an equivariant mapping, and it is not surprising that the well-known
theorem of A. Dold is a very useful tool. Recall that the result claims that there
is no G-equivariant mapping f : X → Y , if G is a finite nontrivial group acting
freely on a cell G-complex Y of dimension at most n, and X is an n-connected
G-space (see [12, 6.2.6]). Volovikov in [15] established a generalization of Dold’s
theorem by relaxing the condition that X is n-connected. Suppose that the group
G = Zp × · · · × Zp is a product of finitely many copies of Zp , with p prime. Let
X and Y be fixed-point free G-spaces such that H̃ i (X; Zp ) = 0 for all i 6 n, and
Y is finite-dimensional and an n-dimensional cohomology sphere over Zp . Then
Volovikov’s theorem says that there is no G-equivariant mapping f : X → Y . In
this paper we include a very similar generalization of Dold’s theorem, with a slightly
simpler proof.
Like Dold’s standard theorem, this result can be easily applied to problems of
topological combinatorics. One of them is Knaster’s problem (from [9]) of finding
all configurations of points A1 , . . . , Ak ∈ S n−1 , such that for every continuous
mapping f : S n−1 → Rm , one can find a rotation ρ of the sphere, such that
f (ρ(A1 )) = · · · = f (ρ(Ak )) (k, n and m are fixed). The configurations of points
2010 Mathematics Subject Classification: Primary 52A35; Secondary 55N91, 05E18, 55M20.
Key words and phrases: G-equivariant mapping, Dold’s theorem, cohomological index,
Knaster’s problem, configuration space, Stiefel manifold.
Supported by the Ministry for Science and Technology of Serbia, Grant 174034.
Communicated by Rade Živaljević.
43
44
JELIĆ
which have some kind of symmetry are interesting because they may admit some
nice actions of groups. Yang established the case k = 3, m = n − 2 of this problem
for vertices of an equilateral triangle (see [18]), and an alternative proof is given
in [16] using the comparison theorem for spectral sequences. Here we show how
the generalization of Dold’s theorem could be used to provide a new proof, much
shorter and simpler than the previous two.
We also consider two cases of Knaster’s problem, studied by Makeev, where
the points on the sphere form a regular polygon or a regular simplex. In particular
for one of the results from [11] we provide here an alternative, short proof. The
other case is proved in [10] by using Dold’s theorem and here we show how the
generalization can improve the result, in the sense that the condition of the theorem
can be weakened. It seems plausible that the other cases of Knaster’s problem,
originally established with the use of Dold’s theorem, admit a generalization, which
can be deduced from the theorem of Volovikov.
2. A generalization of Dold’s theorem
This section contains the generalization of Dold’s theorem which we need for
Knaster’s problem. As we have mentioned, Volovikov in [15] obtained a similar and
even more general result. For the reader’s convenience here we include a slightly
simplified proof where we rely on an elementary dimension argument. As usual, we
consider actions of nontrivial Abelian groups throughout this paper.
Theorem 2.1. Let G be a finite group acting freely on a cell G-complex Y of
dimension at most n, and let X be a G-space. Let R be a commutative ring with
unit such that H n+1 (BG; R) 6= 0 and H̃ k (X; R) = 0 for 0 6 k 6 n. Then there is
no G-equivariant mapping f : X → Y .
Proof. Suppose to the contrary that there exists a G-equivariant mapping
f : X → Y . Let us consider the Fadell-Husseini cohomological indexes of these
spaces, IndG (X) and IndG (Y ) (see [5]). From the monotonicity property of the
index, we have IndG (X) ⊃ IndG (Y ). The idea is to prove the following two facts:
(2.1)
H n+1 (BG; R) * IndG (X), and H n+1 (BG; R) ⊆ IndG (Y ),
which will lead to a contradiction.
We know that IndG (X) = ker p∗X , where pX : XG → BG (XG = EG ×G X),
so p∗X : H ∗ (BG; R) → H ∗ (XG ; R). Our intention is to find index in dimension
∗
n+1
n + 1, i.e. Indn+1
(BG; R) → H n+1 (XG ; R)). Consider the
G (X) = ker(pX : H
cohomology Leray-Serre spectral sequence {Er∗,∗ , dr } [13, Th. 5.2] of the Borel
pX
fibration X → XG −→ BG. It converges to H ∗ (XG ; R) and has the property:
E2p,q ∼
= H p (BG; Hq (X; R)), where Hq (X; R) are local coefficients.
Look at the E2 -term. From the conditions, X is a path connected space,
so we have H p (BG; H0 (X; R)) = H p (BG; H 0 (X; R)) = H p (BG; R), and then
E2p,0 = H p (BG; R), for all p. Above the bottom row, there are n trivial rows.
Indeed, for 1 6 q 6 n, H q (X; R) = 0, so we have simple and trivial coefficients.
ON KNASTER’S PROBLEM
45
We know that homomorphism p∗X is the following composition[13, Th. 5.9]:
n+1,0
H n+1 (BG; R) = E2n+1,0 ։ E3n+1,0 ։ · · · ։ En+1
n+1,0
n+1,0
։ En+2
= E∞
⊂ H n+1 (XG ; R).
But all epimorphisms in this relation are isomorphisms because all differentials dr :
Ern+1−r,r−1 → Ern+1,0 are trivial. Therefore, p∗X is a monomorphism in dimension
n+1
n + 1, and Indn+1
(BG; R) 6= 0, we have H n+1 (BG; R) *
G (X) = 0. Since H
IndG (X), so the first relation in (2.1) is proved.
The remaining part is easy. Since G acts on Y freely, IndG (Y ) = ker p∗Y , where
∗
pY : H ∗ (BG; R) → H ∗ (Y /G; R) [5, 3.15]. But H k (Y /G; R) = 0 for all k > n
because dim Y 6 n. It follows that H k (BG; R) ⊆ IndG (Y ), for all k > n + 1. Thus
(2.1) is proved. Finally, from (2.1) and the monotonicity of index, we obtain a
contradiction which proves the theorem.
3. Knaster’s problem
Now we apply the theorem from previous section to Knaster’s problem. Consider a finite set of points A = {A1 , A2 , . . . , Ak } on the standard sphere S n−1 ,
and fix some m ∈ N. Knaster’s problem asks, for a given arbitrary continuous
mapping f : S n−1 → Rm , whether there exists a rotation ρ ∈ SO(n), such that
f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Ak )). The problem was originally stated only for
k = n − m + 1 (it can be shown that for a positive answer, A should lie in some
(n − m)-plane). The general answer to the problem is negative; counterexamples
were found by Makeev in [10], and later by Babenko and Bogatyi in [1], and then
by Chen in [4].
However, for many “nice" configurations of A, the answer to the problem is
affirmative. The most famous case is m = n − 1, k = 2, A = {e1 , −e1 }, which is
the Borsuk–Ulam theorem: for every continuous function f : S n−1 → Rn−1 , there
exists an x ∈ S n−1 such that f (x) = f (−x). This case was generalized by Hopf
[8] for any two-element subset of S n−1 . There are also many positive answers for
m = 1: when A is any 3-element subset of S 2 [6]; A = {e1 , e2 , . . . , en } ⊂ S n−1
being the standard orthonormal basis [17]; A ⊂ S n−1 being the set of vertices of
any regular (n − 1)-simplex [2], etc.
In [18], Yang proved the positive result for the case when m = n − 2 and A
is the vertex set of an equilateral triangle on a great circle of S n−1 (a great circle
of sphere is any circle on the sphere with center at the origin). Here we give much
shorter proof for that case, using Theorem 2.1.
Theorem 3.1. Let A1 , A2 , A3 ∈ S n−1 be the vertices of an equilateral triangle
on a great circle (n > 2). Then for every continuous mapping f : S n−1 → Rn−2 ,
there exists a rotation ρ ∈ SO(n) such that f (ρ(A1 )) = f (ρ(A2 )) = f (ρ(A3 )).
Proof. Consider the configuration space of all triples of points (X1 , X2 , X3 )
that are vertices of some equilateral triangle on some great circle. It can be viewed
as Stiefel manifold V2 (Rn ) (the third point X3 is uniquely determined by the first
two, and (X1 , X2 ) is any pair of points that make constant angle 2π
3 ).
46
JELIĆ
Suppose to the contrary that there is a continuous mapping f : S n−1 → Rn−2 ,
for which a desired rotation ρ doesn’t exist. Since the vertex set of any considered
triangle can be obtained from the set {A1 , A2 , A3 } by some rotation, it means that
there does not exist equilateral triangle on any great circle whose vertices have the
same image under f . Define a mapping: F : V2 (Rn ) → Rn−2 × Rn−2 × Rn−2 by
F (X1 , X2 , X3 ) := (f (X1 ), f (X2 ), f (X3 )).
By assumption, F (V2 (Rn )) has empty intersection with the diagonal ∆ in the space
Rn−2 × Rn−2 × Rn−2 (dim ∆ = n − 2). So we can take smaller codomain and
have F : V2 (Rn ) → (Rn−2 )3 r ∆. There are natural Z3 -actions on these spaces.
The action of Z3 cyclically permutes vectors of V2 (Rn ) (with our identification
of configuration space with V2 (Rn ), it means that the generator of Z3 acts as:
gZ3 (X1 , X2 , X3 ) := (X2 , X3 , X1 )). Also, Z3 cyclically permutes (n − 2)-dimensional
vectors in (Rn−2 )3 r ∆. Each of these actions are free and F is a Z3 -equivariant
mapping.
Now take the orthogonal projection p : (Rn−2 )3 r ∆ → ∆⊥ r {0}, and radial
projection r from ∆⊥ r {0} onto the unit sphere in ∆⊥ , which is S 2n−5 . Spaces
∆⊥ r {0} and S 2n−5 have inherited Z3 -actions from (Rn−2 )3 r ∆, and it is obvious
that p and r are Z3 -equivariant deformations. So we have the following composition
which is a Z3 -equivariant mapping, and Z3 -actions on the spaces are free:
Z
3
φ = r ◦ p ◦ F : V2 (Rn ) −→
S 2n−5 .
It remains to show that such equivariant mapping doesn’t exist. The domain is
(n − 3)-connected [7, 4.53], while the codomain has larger dimension, so we cannot
apply Dold’s standard theorem. But we can apply Theorem 2.1, for which we need
cohomology groups of V2 (Rn ). If n is odd, H k (V2 (Rn ); Z) ∼
= Z for k ∈ {0, 2n − 3},
H n−1 (V2 (Rn ); Z) ∼
= Z2 , and other cohomology groups are trivial. If n is even,
H k (V2 (Rn ); Z) ∼
= H k (S n−1 × S n−2 ; Z) [3, Prop. 10.1]. We have two cases.
◦
1 Let n be odd. From the universal coefficient theorem we derive cohomology
groups with Z3 -coefficients: H k (V2 (Rn ); Z3 ) ∼
= Z3 for k ∈ {0, 2n − 3}, and it is
trivial otherwise. Since H k (BZ3 ; Z3 ) 6= 0 for all k [14, III 2.5], Theorem 2.1 can be
applied to spaces V2 (Rn ) and S 2n−5 , and G = R = Z3 . It implies that there is no
Z
3
equivariant mapping V2 (Rn ) −→
S 2n−5 , which is a contradiction. Thus theorem is
proved in this case.
2◦ Let n be even. Consider the mapping g : S n → Rn−1 , g := Sf -the suspension of f (the image of f is compact so the codomain of suspension is a subspace of
Rn−1 ). Suppose that there are three points on S n that are vertices of a great equilateral triangle having the same image under g. From the definition of suspension,
those points have to be on the same “level" S n−1 × {t} of the suspension S n . Moreover, in order to be on a great circle, they have to be on the middle level S n−1 ×{0}.
Then we obtain the vertices of a great equilateral triangle of sphere S n−1 with the
same image by f , which is impossible. So, those points cannot exist. Then function
Z3
g, in the same way as f dues, induces the mapping G : V2,n+1 −→
(Rn−1 )3 r ∆.
◦
Since n + 1 is an odd integer, we obtain a contradiction from 1 .
ON KNASTER’S PROBLEM
47
The following theorem, which is stated in Makeev’s paper [11], can be proved
in the completely same way for n odd.
Theorem 3.2. Let p be an odd prime, n, m ∈ N, n odd, such that (p − 1) · m <
2n − 2. Let A1 , A2 , . . . , Ap be the vertices of a regular polygon with p sides, on a
great circle of sphere S n−1 . Then for every continuous mapping f : S n−1 → Rm ,
there exists a rotation ρ ∈ SO(n) such that f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Ap )).
The configuration space is again V2 (Rn ) because every regular polygon on a
great circle, with p sides, is determined by the first two vertices. Then we consider
the mapping F : V2 (Rn ) → Rm × · · · × Rm , F (X1 , . . . , Xp ) := (f (X1 ), . . . , f (Xp )),
{z
}
|
p
where we suppose that f is a mapping that contradicts the theorem. Like in the
previous theorem, we would obtain a sequence of Zp -equivariant mappings:
Zp
Zp
Zp
V2 (Rn ) −→ (Rm )p r ∆ −→ ∆⊥ r {0} −→ S m(p−1)−1 .
Since n is odd, H k (V2 (Rn ); Zp ) ∼
= Zp for k ∈ {0, 2n − 3}, and it is trivial otherwise.
Also, H k (BZp ; Zp ) isn’t trivial [14, III 2.5]. We have 2n − 4 > m(p − 1) − 1, so
Theorem 2.1. gives a contradiction again and proves the theorem.
Let us show one more application of Theorem 2.1. By using Dold’s standard
theorem, Makeev proved the following theorem in [10].
Theorem 3.3 (Makeev). Let p be an odd prime and n ∈ N, such that 2p < n+1.
Let A1 , A2 , . . . , Ap ∈ S n−1 be the vertices of a regular (p − 1)-simplex. Then for
every continuous mapping f : S n−1 → R, there exists a rotation ρ ∈ SO(n) such
that f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Ap )).
With the generalization of Dold’s theorem, we are able to prove Makeev’s result
with a weaker condition: 3p 6 2n + 2, for n even. (For n odd, our technique gives
the estimate from Makeev’s theorem.) The proof follows the idea of Makeev’s proof.
Theorem 3.4. Let p be an odd prime, n ∈ N, n even, such that 3p 6 2n + 2.
Let A1 , A2 , . . . , Ap ∈ S n−1 be the vertices of a regular (p − 1)-simplex, whose centre
is not at the origin. Then for every continuous mapping f : S n−1 → R, there exists
a rotation ρ ∈ SO(n) such that f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Ap )).
Proof. Suppose that f is a continuous mapping for which there is no desired
rotation. The configuration space of all p-tuples (X1 , . . . , Xp ) which form a regular
(p−1)-simplex congruent with simplex (A1 , . . . , Ap ), is the Stiefel manifold Vp (Rn ).
(If A1 , . . . , Ap are vertices of an orthonormal frame, it is obvious. Else, every
(X1 , . . . , Xp ) has its unique corresponding simplex (X1′ , . . . , Xp′ ) formed by vertices
of an orthonormal frame, and whose centre is collinear with the origin and the centre
of (X1 , . . . , Xp ).) Further, since vertex set {X1 , . . . , Xp } of every considered (p−1)simplex is ρ({A1 , . . . , Ap }), for some ρ ∈ SO(n), then X1 , . . . , Xp never have the
same image by f . So the following mapping is well defined: F : Vp (Rn ) → Rp r ∆,
F (X1 , . . . , Xp ) := (f (X1 ), . . . , f (Xp )). As in Theorem 3.1, now we have reductions
by orthogonal and radial projection:
Vp (Rn ) → Rp r ∆ → ∆⊥ r {0} → S p−2 .
48
JELIĆ
Group Zp naturally acts on these spaces, on the first two by cyclically permuting
the coordinates, and on ∆⊥ r{0} and S p−2 actions are inherited from Rp . All these
actions are free and all mappings are Zp -equivariant. It remains to show that there
is no Zp -equivariant mapping Vp (Rn ) → S p−2 . Makeev got the contradiction from
Dold’s standard theorem for 2p < n + 1. For a stronger result, let us consider the
cohomology H ∗ (Vp (Rn ); Zp ). It is isomorphic to the cohomological algebra (with
Zp -coefficients) of the product S 2n−3 × S 2n−7 × · · · × S 2q+1 , multiplied with S n−1
if 2|n, and multiplied with S q if 2 | q, where q denotes n − p, q is the smallest odd
integer which is > q and n is the largest odd integer which is 6 n [3, Prop. 10.2].
In our case, 2 | n, 2 ∤ p, so 2 ∤ q = q, and we have H ∗ (Vp (Rn ); Zp ) ∼
= H ∗ (X; Zp ),
e k (X; Zp ) = 0, for all
where X = S 2n−5 × S 2n−9 × · · · × S 2(n−p)+1 × S n−1 . Then H
k 6 min{2(n− p), n− 2}. From 3p 6 2n+ 2 we conclude that min{2(n− p), n− 2} >
p − 2 = dim S p−2 . Then Theorem 2.1 (for G = R = Zp ) implies nonexistence of
Zp -equivariant mapping Vp (Rn ) → S p−2 , which is a contradiction.
There is an analogous result when the codomain is Rm , m > 2.
Theorem 3.5. Let p be an odd prime, n, m ∈ N, n even and m > 2 such
that (p − 1)m + 1 6 n. Let A1 , A2 , . . . , Ap ∈ S n−1 be the vertices of a regular
simplex. Then for every continuous mapping f : S n−1 → Rm , there exists a rotation
ρ ∈ SO(n) such that f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Ap )).
The proof completely follows from the previous one, so we omit it. We would
construct a Zp -equivariant mapping Vp (Rn ) → S (p−1)m−1 , and obtain a contradiction for min{2(n − p), n − 2} > (p − 1)m − 1, which is equivalent to the condition
(p − 1)m + 1 6 n, for m > 2.
The technique applied throughout this paper provides the following similar
result for the vertex set of a regular (pk − 1)-simplex.
Theorem 3.6. Let p be a prime, n, k ∈ N, k > 1. Let A1 , A2 , . . . , Apk ∈ S n−1
be the vertices of a regular (pk − 1)-simplex, whose center is not at the origin. If
numbers p, k and n satisfy one of the following conditions:
• p is odd, n is even and 2n + 2 > 3pk ,
• p is odd, n is odd and n + 1 > 2pk ,
• p = 2, n is even and n + 1 > 2k+1 ,
• p = 2, n is odd and 2n + 2 > 3 · 2k ,
then for every continuous mapping f : S n−1 → R, there exists a rotation ρ ∈ SO(n)
such that f (ρ(A1 )) = f (ρ(A2 )) = · · · = f (ρ(Apk )).
The proof is analogous to the proof of Theorem 3.4, so it is omitted. We would
make obvious modifications (consider the actions of group (Zp )k instead of Zp ),
and an application of Volovikov’s theorem [15], instead of Theorem 2.1, implies a
contradiction.
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Department of Topology
Faculty of Mathematics
University of Belgrade
Belgrade
Serbia
marijaj@matf.bg.ac.rs
(Received 28 04 2015)
(Revised 30 10 2015)